Answered: Lorentz contraction In relativity…

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Lorentz contraction In relativity theory, the length of an object, say a rocket, appears to an observer to depend on the speed at which the object is traveling with respect to the observer. If the observer measures the rocket’s length as L0 at rest, then at speed y the length will appear to be

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Step 1

Length contraction is one of the most interesting phenomena that are explained by this theory. other interesting phenomena are time dilation, Relative speed, etc.

Length contraction:

If we measure the length of any object that is moving relative to our frame, we observe that its length $L$ gets smaller than the proper length $L_{o}$ that would have been measured if the object is at rest. At very high speeds i.e. closer to the speed of light, the lengths or distances measured by different observers are different.

This phenomenon is also known as Lorentz Contraction. It $L_{o}$ is the length of the object when observed from the rest frame then the length of the object ( $L$ ) measured in a frame moving with velocity $v$ is given as:

$L = L_{o} \sqrt{1 - \frac{v^{2}}{c^{2}}}$ , $c$ is the speed of light

Step 2

We know the formula for Length contraction is given as: $L = L_{o} \sqrt{1 - \frac{v^{2}}{c^{2}}}$

As $v$ increases then the value $\frac{v^{2}}{c^{2}}$ increases and $- \frac{v^{2}}{c^{2}}$ decreases. Hence, the quantity $1 - \frac{v^{2}}{c^{2}}$ decreases which eventually decreases the length of the object that is moving with velocity $v$ .

(i) $v \leq c$

If $v \leq c$ then, $\frac{v}{c} \leq 1$ .

This implies that $\frac{v^{2}}{c^{2}} \leq 1$ and that gives $0 \leq 1 - \frac{v^{2}}{c^{2}}$

Thus, the expression $\sqrt{1 - \frac{v^{2}}{c^{2}}}$ is defined for this case.

(ii) $v > c$

If $v > c$ then $\frac{v}{c} > 1$

This implies that $\frac{v^{2}}{c^{2}} > 1$ that gives $1 - \frac{v^{2}}{c^{2}} < 0$

Thus, the expression $\sqrt{1 - \frac{v^{2}}{c^{2}}}$ is not defined for this case as the term inside the square root is negative which will give the result in the complex domain. Therefore, only the left-hand limit is considered while approaching from $v t o c$ .

The Lorentz contraction is given by : $L = L_{o} \sqrt{1 - \frac{v^{2}}{c^{2}}}$

Now, applying the Left-hand limit i.e. $v \to c$ on both sides,

$\lim_{v \to c} L = \lim_{v \to c} L_{o} \sqrt{1 - \frac{v^{2}}{c^{2}}} \lim_{v \to c} L = L_{o} \lim_{v \to c} \sqrt{1 - \frac{v^{2}}{c^{2}}} = L_{o} \sqrt{1 - 1} = L_{o} \times 0 = 0$

From above we observe that if the object moves with a velocity equal to the speed of light the length that is observed will be zero which is not the case for practical purposes. Hence, the case $v = c$ gives the contracted length to be zero. Thus we do not consider the case for $v > c$ i.e. Right-hand limit as the values of contracted length will become complex. Thus only the Left-hand limit is considered.

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